Calculus_Volume_1_-_WEB_68M1Z5W-62

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x f(x) f′(x) g(x) g′(x)
0 2 5 0 2
1 1 −2 3 0
2 4 4 1 −1
3 3 −3 2 3
245. h(x) = f ⎛⎝g(x)⎞⎠; a = 0
246. h(x) = g⎛⎝ f (x)⎞⎠; a = 0
247. h(x) = ⎛⎝x4 + g(x)⎞⎠
−2
; a = 1
248. h(x) = ⎛⎝
f (x)
g(x)
⎞
⎠
2
; a = 3
249. h(x) = f ⎛⎝x + f (x)⎞⎠; a = 1
250. h(x) = ⎛
⎝1 + g(x)⎞⎠3; a = 2
251. h(x) = g⎛⎝2 + f ⎛⎝x2⎞⎠
⎞
⎠; a = 1
252. h(x) = f ⎛⎝g(sinx)⎞⎠; a = 0
253. [T] The position function of a freight train is given by
s(t) = 100(t + 1)−2, with s in meters and t in seconds.
At time t = 6 s, find the train’s
a. velocity and
b. acceleration.
c. Using a. and b. is the train speeding up or slowing
down?
254. [T] A mass hanging from a vertical spring is in
simple harmonic motion as given by the following position
function, where t is measured in seconds and s is in
inches: s(t) = −3cos⎛⎝πt + π
4
⎞
⎠.
a. Determine the position of the spring at t = 1.5 s.
b. Find the velocity of the spring at t = 1.5 s.
255. [T] The total cost to produce x boxes of Thin Mint
Girl Scout cookies is C dollars, where
C = 0.0001x3 − 0.02x2 + 3x + 300. In t weeks
production is estimated to be x = 1600 + 100t boxes.
a. Find the marginal cost C′ (x).
b. Use Leibniz’s notation for the chain rule,
dC
dt = dC
dx · dxdt , to find the rate with respect to
time t that the cost is changing.
c. Use b. to determine how fast costs are increasing
when t = 2 weeks. Include units with the answer.
256. [T] The formula for the area of a circle is A = πr2,
where r is the radius of the circle. Suppose a circle is
expanding, meaning that both the area A and the radius r
(in inches) are expanding.
a. Suppose r = 2 − 100
(t + 7)2 where t is time in
seconds. Use the chain rule dA
dt = dA
dr · drdt to find
the rate at which the area is expanding.
b. Use a. to find the rate at which the area is
expanding at t = 4 s.
257. [T] The formula for the volume of a sphere is
S = 4
3πr
3, where r (in feet) is the radius of the sphere.
Suppose a spherical snowball is melting in the sun.
a. Suppose r = 1
(t + 1)2 − 1
12 where t is time in
minutes. Use the chain rule dS
dt = dS
dr · drdt to find
the rate at which the snowball is melting.
b. Use a. to find the rate at which the volume is
changing at t = 1 min.
258. [T] The daily temperature in degrees Fahrenheit of
Phoenix in the summer can be modeled by the function
T(x) = 94 − 10cos⎡⎣ π12(x − 2)⎤⎦, where x is hours after
midnight. Find the rate at which the temperature is
changing at 4 p.m.
259. [T] The depth (in feet) of water at a dock changes
with the rise and fall of tides. The depth is modeled by
the function D(t) = 5sin⎛⎝π6 t − 7π
6
⎞
⎠+ 8, where t is the
number of hours after midnight. Find the rate at which the
depth is changing at 6 a.m.
298 Chapter 3 | Derivatives
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3.7 | Derivatives of Inverse Functions
Learning Objectives
3.7.1 Calculate the derivative of an inverse function.
3.7.2 Recognize the derivatives of the standard inverse trigonometric functions.
In this section we explore the relationship between the derivative of a function and the derivative of its inverse. For functions
whose derivatives we already know, we can use this relationship to find derivatives of inverses without having to use the
limit definition of the derivative. In particular, we will apply the formula for derivatives of inverse functions to trigonometric
functions. This formula may also be used to extend the power rule to rational exponents.
The Derivative of an Inverse Function
We begin by considering a function and its inverse. If f (x) is both invertible and differentiable, it seems reasonable that
the inverse of f (x) is also differentiable. Figure 3.28 shows the relationship between a function f (x) and its inverse
f −1 (x). Look at the point ⎛⎝a, f −1 (a)⎞⎠ on the graph of f −1(x) having a tangent line with a slope of ⎛⎝ f −1⎞⎠′ (a) = p
q . This
point corresponds to a point ⎛⎝ f −1 (a), a⎞⎠ on the graph of f (x) having a tangent line with a slope of f ′ ⎛⎝ f −1 (a)⎞⎠ = q
p.
Thus, if f −1(x) is differentiable at a, then it must be the case that
⎛
⎝ f −1⎞⎠′ (a) = 1
f ′ ⎛⎝ f −1 (a)⎞⎠
.
Figure 3.28 The tangent lines of a function and its inverse are
related; so, too, are the derivatives of these functions.
We may also derive the formula for the derivative of the inverse by first recalling that x = f ⎛⎝ f −1 (x)⎞⎠. Then by
differentiating both sides of this equation (using the chain rule on the right), we obtain
1 = f ′ ⎛⎝ f −1 (x)⎞⎠
⎛
⎝ f −1 )′(x)⎞⎠.
Solving for ( f −1 )′(x), we obtain
(3.19)⎛
⎝ f −1⎞⎠′ (x) = 1
f ′ ⎛⎝ f −1 (x)⎞⎠
.
We summarize this result in the following theorem.
Chapter 3 | Derivatives 299
3.42
Theorem 3.11: Inverse Function Theorem
Let f (x) be a function that is both invertible and differentiable. Let y = f −1 (x) be the inverse of f (x). For all x
satisfying f ′ ⎛⎝ f −1 (x)⎞⎠ ≠ 0,
dy
dx = d
dx
⎛
⎝ f −1(x)⎞⎠ = ⎛⎝ f −1⎞⎠′ (x) = 1
f ′ ⎛⎝ f −1 (x)⎞⎠
.
Alternatively, if y = g(x) is the inverse of f (x), then
g '(x) = 1
f ′ ⎛⎝g(x)⎞⎠
.
Example 3.60
Applying the Inverse Function Theorem
Use the inverse function theorem to find the derivative of g(x) = x + 2
x . Compare the resulting derivative to that
obtained by differentiating the function directly.
Solution
The inverse of g(x) = x + 2
x is f (x) = 2
x − 1. Since g′ (x) = 1
f ′ ⎛⎝g(x)⎞⎠
, begin by finding f ′ (x). Thus,
f ′ (x) = −2
(x − 1)2 and f ′ ⎛⎝g(x)⎞⎠ = −2
⎛
⎝g(x) − 1⎞⎠2
= −2
⎛
⎝x + 2
x − 1⎞⎠
2 = − x2
2 .
Finally,
g′ (x) = 1
f ′ ⎛⎝g(x)⎞⎠
= − 2
x2.
We can verify that this is the correct derivative by applying the quotient rule to g(x) to obtain
g′ (x) = − 2
x2.
Use the inverse function theorem to find the derivative of g(x) = 1
x + 2. Compare the result obtained
by differentiating g(x) directly.
Example 3.61
Applying the Inverse Function Theorem
Use the inverse function theorem to find the derivative of g(x) = x3 .
300 Chapter 3 | Derivatives
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3.43
Solution
The function g(x) = x3 is the inverse of the function f (x) = x3. Since g′ (x) = 1
f ′ ⎛⎝g(x)⎞⎠
, begin by finding
f ′ (x). Thus,
f ′ (x) = 3x2 and f ′ ⎛⎝g(x)⎞⎠ = 3⎛⎝ x3 ⎞⎠
2
= 3x2/3.
Finally,
g′ (x) = 1
3x2/3 = 1
3x
−2/3.
Find the derivative of g(x) = x5 by applying the inverse function theorem.
From the previous example, we see that we can use the inverse function theorem to extend the power rule to exponents of
the form 1
n, where n is a positive integer. This extension will ultimately allow us to differentiate xq, where q is any
rational number.
Theorem 3.12: Extending the Power Rule to Rational Exponents
The power rule may be extended to rational exponents. That is, if n is a positive integer, then
(3.20)d
dx
⎛
⎝x1/n⎞⎠ = 1
nx
(1/n) − 1.
Also, if n is a positive integer and m is an arbitrary integer, then
(3.21)d
dx
⎛
⎝xm/n⎞⎠ = m
n x
(m/n) − 1.
Proof
The function g(x) = x1/n is the inverse of the function f (x) = xn. Since g′ (x) = 1
f ′ ⎛⎝g(x)⎞⎠
, begin by finding f ′ (x).
Thus,
f ′ (x) = nxn − 1 and f ′ ⎛⎝g(x)⎞⎠ = n(x1/n)n − 1 = nx(n − 1)/n.
Finally,
g′ (x) = 1
nx(n − 1)/n = 1
nx
(1 − n)/n = 1
nx
(1/n) − 1.
To differentiate xm/n we must rewrite it as ⎛⎝x1/n⎞⎠
m
and apply the chain rule. Thus,
d
dx
⎛
⎝xm/n⎞⎠ = d
dx
⎛
⎝⎛⎝x1/n⎞⎠
m⎞
⎠ = m⎛⎝x1/n⎞⎠
m − 1
· 1
nx
(1/n) − 1 = m
n x
(m/n) − 1.
□
Chapter 3 | Derivatives 301
3.44
Example 3.62
Applying the Power Rule to a Rational Power
Find the equation of the line tangent to the graph of y = x2/3 at x = 8.
Solution
First find
dy
dx and evaluate it at x = 8. Since
dy
dx = 2
3x
−1/3 and dy
dx |x = 8
= 1
3
the slope of the tangent line to the graph at x = 8 is 1
3.
Substituting x = 8 into the original function, we obtain y = 4. Thus, the tangent line passes through the point
(8, 4). Substituting into the point-slope formula for a line, we obtain the tangent line
y = 1
3x + 4
3.
Find the derivative of s(t) = 2t + 1.
Derivatives of InverseTrigonometric Functions
We now turn our attention to finding derivatives of inverse trigonometric functions. These derivatives will prove invaluable
in the study of integration later in this text. The derivatives of inverse trigonometric functions are quite surprising in that
their derivatives are actually algebraic functions. Previously, derivatives of algebraic functions have proven to be algebraic
functions and derivatives of trigonometric functions have been shown to be trigonometric functions. Here, for the first time,
we see that the derivative of a function need not be of the same type as the original function.
Example 3.63
Derivative of the Inverse Sine Function
Use the inverse function theorem to find the derivative of g(x) = sin−1 x.
Solution
Since for x in the interval
⎡
⎣−π
2, π2
⎤
⎦, f (x) = sinx is the inverse of g(x) = sin−1 x, begin by finding f ′(x).
Since
f ′ (x) = cosx and f ′ ⎛⎝g(x)⎞⎠ = cos ⎛⎝sin−1 x⎞⎠ = 1 − x2,
we see that
g′ (x) = d
dx
⎛
⎝sin−1 x⎞⎠ = 1
f ′ ⎛⎝g(x)⎞⎠
= 1
1 − x2
.
302 Chapter 3 | Derivatives
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	Chapter 3. Derivatives
	3.7. Derivatives of Inverse Functions*